Simplify the following expression and state the condition under which the simplification is valid. You can assume that $a \neq 0$. $k = \dfrac{9(3a + 2)}{4} \div \dfrac{15a^2 + 10a}{5a} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{9(3a + 2)}{4} \times \dfrac{5a}{15a^2 + 10a} $ When multiplying fractions, we multiply the numerators and the denominators. $k = \dfrac{ 9(3a + 2) \times 5a } { 4 \times (15a^2 + 10a) } $ $ k = \dfrac {5a \times 9(3a + 2)} {4 \times 5a(3a + 2)} $ $ k = \dfrac{45a(3a + 2)}{20a(3a + 2)} $ We can cancel the $3a + 2$ so long as $3a + 2 \neq 0$ Therefore $a \neq -\dfrac{2}{3}$ $k = \dfrac{45a \cancel{(3a + 2})}{20a \cancel{(3a + 2)}} = \dfrac{45a}{20a} = \dfrac{9}{4} $